(b^2+3b)+b+3=0 — step-by-step solution

Let's solve in

ℝ

b^{2}

+ 3 b

+ b

+ 3 = 0

ⓘ Like terms are grouped by adding their coefficients:

3 b + b

4 b

b^{2} + 4 b + 3 = 0

Given the quadratic form

A b^{2} + B b + C

, you can calculate the discriminant:

Δ = B^{2} - 4 A C

. Here

A = 1

B = 4

and

C = 3

Δ = 4^{2} - 4 \cdot 1 \cdot 3

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

4^{2}

4 \cdot 4

16

Δ = 16 - 4 \cdot 3

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 16 - 12

ⓘ Start from

16

and move back

12

on the number line:

16 - 12

4

Δ = 4

Δ > 0

, the equation has two roots:

The quadratic formula gives

b_{1}

\frac{- B - \sqrt{Δ}}{2 A}

\frac{- 4 - \sqrt{4}}{2 \cdot 1}

b_{1} = \frac{- 4 - \sqrt{4}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{4}

2

;

2^{2}

4

b_{1} = \frac{- 4 - 2}{2}

ⓘ The numerator and the denominator are computed before the division:

- 4 - 2

-6

b_{1} = - \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{6}{2}

-3

b_{1} = -3

The quadratic formula gives

b_{2}

\frac{- B + \sqrt{Δ}}{2 A}

\frac{- 4 + \sqrt{4}}{2 \cdot 1}

b_{2} = \frac{- 4 + \sqrt{4}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{4}

2

;

2^{2}

4

b_{2} = \frac{- 4 + 2}{2}

ⓘ The numerator and the denominator are computed before the division:

- 4 + 2

-2

b_{2} = - \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{2}

-1

b_{2} = -1

▶Solutions found:

b = {-3, -1}